# Does projective duality preserve arithmetic-Cohen-Macaulay-ness?

Let $V$ be a vector space over $\mathbb{C}$. Suppose $X\subset \mathbb{P} V$ is an algebraic variety, and consider its projective dual variety $X^\vee \subset \mathbb{P} V^*$. If the coordinate ring $\mathbb{C}[X]$ is arithmetically Cohen-Macaulay (aCM) is the same true for $\mathbb{C}[X^\vee]$?

An example where this is true is when $X$ is $n\times n$ matrices of rank $1$ and $X^\vee$ is $n\times n$ matrices of rank $n-1$.

I guess that it is true in general. Does someone have a reference (or a counter-example)?

Thanks!

• I guess I should rephrase the question: What reasonable conditions on $X$ will guarantee that $X^\vee$ is aCM? Aug 25 '14 at 16:19

Not true. Start from a variety $Y\subset \mathbb{P}V^*$ which is not aCM -- for instance a quartic rational curve in $\mathbb{P}^3$. In most cases, in particular in the example, its dual $X:=Y^\vee$ is a hypersurface in $\mathbb{P}V$, hence is aCM. But $X^\vee=Y$ is not.
• And is there an example with $X$ smooth? Aug 21 '14 at 8:33